Electromagnetic Waves — Page 6 | JEE Physics

Q51

The rms value of conduction current in a parallel plate capacitor is

6.9 \,\mu \mathrm{A}

. The capacity of this capacitor, if it is connected to

230 \mathrm{~V}

ac supply with an angular frequency of

600 \,\mathrm{rad} / \mathrm{s}

, will be :

A 5 pF

B 50 pF

C 100 pF

D 200 pF

Correct Answer

Option B

Solution

{Z_C} = {V \over I}

\Rightarrow {1 \over {\omega C}} = {{230} \over {6.9}}\,M\,\Omega

\Rightarrow C = {{6.9} \over {230\,\omega }}\,\mu F

= {{6.9} \over {230 \times 600}}\,\mu F

C = 50\,pF

Q52

Light wave travelling in air along x-direction is given by

{E_y} = 540\sin \pi \times {10^4}(x - ct)\,V{m^{ - 1}}

. Then, the peak value of magnetic field of wave will be (Given c = 3

\times

108 ms

-

1)

A 18

\times

10

-

7 T

B 54

\times

10

-

7 T

C 54

\times

10

-

8 T

D 18

\times

10

-

8 T

Correct Answer

Option A

Solution

c = 3 \times {10^8}

m/sec

B = {E \over c} = {{540} \over {3 \times {{10}^8}}} = 18 \times {10^{ - 7}}\,T

Q53

The magnetic field of a plane electromagnetic wave is given by :

\overrightarrow{\mathrm{B}}=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \,\hat{j} \mathrm{~T}

. The amplitude of the electric field would be :

A

6\, \mathrm{Vm}^{-1}

along

x

-axis

B

3\, \mathrm{Vm}^{-1}

along

z

-axis

C

6\, \mathrm{Vm}^{-1}

along

z

-axis

D

2 \times 10^{-8} \,\mathrm{Vm}^{-1}

along

z

-axis

Correct Answer

Option C

Solution

Speed of light

c = {\omega \over k} = {{1.5 \times {{10}^{11}}} \over {0.5 \times {{10}^3}}} = 3 \times {10^8}

m/sec So,

{E_0} = {B_0}c

= 2 \times {10^{ - 8}} \times 3 \times {10^8}

= 6

V/m Direction will be along z-axis.

Q54

The oscillating magnetic field in a plane electromagnetic wave is given by

B_{y}=5 \times 10^{-6} \sin 1000 \pi\left(5 x-4 \times 10^{8} t\right) T

. The amplitude of electric field will be :

A

15 \times 10^{2} \,\mathrm{Vm}^{-1}

B

5 \times 10^{-6} \,\mathrm{Vm}^{-1}

C

16 \times 10^{12} \,\mathrm{Vm}^{-1}

D

4 \times 10^{2} \,\mathrm{Vm}^{-1}

Correct Answer

Option D

Solution

In an electromagnetic wave, the magnitude of the electric field (E) and the magnetic field (B) are related by the speed of the wave (v), which can be represented as :

E = Bv

Here, B is the magnetic field, E is the electric field, and v is the speed of the electromagnetic wave.

In the given problem, the peak value of the magnetic field is given as :

B_0 = 5 \times 10^{-6} T

The wave speed (v) can be calculated from the given equation for the magnetic field, using the relationship between frequency (f), wave number (k), and speed :

v = \frac{\omega}{k} = \frac{1000 \pi \times 4 \times 10^8}{1000 \pi \times 5} = 8 \times 10^7 m/s

Substituting these values into the equation for the electric field gives the peak value of the electric field :

E_0 = B_0 v = 5 \times 10^{-6} T \times 8 \times 10^7 m/s = 4 \times 10^2 V/m

Q55

A velocity selector consists of electric field

\vec{E}=E \,\hat{k}

and magnetic field

\vec{B}=B \,\hat{j}

with

B=12 \,m T

. The value of

E

required for an electron of energy

728 \,\mathrm{e} V

moving along the positive

x

-axis to pass undeflected is : (Given, mass of electron

=9.1 \times 10^{-31} \mathrm{~kg}

)

A

192 \,\mathrm{kVm}^{-1}

B

192 \,\mathrm{mVm}^{-1}

C

9600 \,\mathrm{kVm}^{-1}

D

16 \,\mathrm{kVm}^{-1}

Correct Answer

Option A

Solution

v = {E \over B}

and

K = {1 \over 2}m{v^2}

\Rightarrow \sqrt {{{2K} \over m}} \times B = E

\Rightarrow E = \sqrt {{{2 \times 728 \times 1.6 \times {{10}^{ - 19}}} \over {9.1 \times {{10}^{ - 31}}}}} \times 12 \times {10^{ - 3}}

= 192000

V/m

Q56

The energy of an electromagnetic wave contained in a small volume oscillates with

A double the frequency of the wave

B the frequency of the wave

C half the frequency of the wave

D zero frequency

Correct Answer

Option A

Solution

The energy of an electromagnetic wave contained in a small volume oscillates with double the frequency of the wave.

Here's why: The electromagnetic wave consists of oscillating electric and magnetic fields.

The energy density of the wave is proportional to the square of the amplitude of these fields.

Since the square of a sinusoidal function oscillates with twice the frequency of the original function, the energy density of the electromagnetic wave also oscillates with twice the frequency of the wave.

Q57

Identify the correct statements from the following descriptions of various properties of electromagnetic waves. (A) In a plane electromagnetic wave electric field and magnetic field must be perpendicular to each other and direction of propagation of wave should be along electric field or magnetic field. (B) The energy in electromagnetic wave is divided equally between electric and magnetic fields. (C) Both electric field and magnetic field are parallel to each other and perpendicular to the direction of propagation of wave. (D) The electric field, magnetic field and direction of propagation of wave must be perpendicular to each other. (E) The ratio of amplitude of magnetic field to the amplitude of electric field is equal to speed of light. Choose the most appropriate answer from the options given below :

A (D) only

B (B) and (D) only

C (B), (C) and (E) only

D (A), (B) and (E) only

Correct Answer

Option B

Solution

In an EM wave : 1.

\overrightarrow E \,\, \bot \,\,\overrightarrow B

2.

\overrightarrow V \equiv \overrightarrow E \times \overrightarrow B

3. Energy is equally divided 4.

\left| {\overrightarrow V } \right| = \left| {\overrightarrow E } \right|/\left| {\overrightarrow B } \right|

Q58

The ratio of average electric energy density and total average energy density of electromagnetic wave is :

A 1

B 3

C 2

D

\dfrac{1}{2}

Correct Answer

Option D

Solution

Avg electric energy density $=\dfrac{1}{4} \varepsilon_0 \mathrm{E}_0^2$ Total Avg energy density $=\dfrac{1}{2} \varepsilon_0 \mathrm{E}_0^2$ $\therefore$ Ratio of average electric energy density and total Avg energy density

= \frac{\frac{1}{4} \varepsilon_0 \mathrm{E}_0^2}{\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2}=\frac{2}{4}=\frac{1}{2}

Q59

A point source of

100 \mathrm{~W}

emits light with

5 \%

efficiency. At a distance of

5 \mathrm{~m}

from the source, the intensity produced by the electric field component is:

A

\dfrac{1}{40 \pi} \dfrac{W}{m^2}

B

\dfrac{1}{10 \pi} \dfrac{W}{m^2}

C

\dfrac{1}{20 \pi} \dfrac{W}{m^2}

D

\dfrac{1}{2 \pi} \dfrac{W}{m^2}

Correct Answer

Option A

Solution

A point source of $100 \mathrm{~W}$ emits light with $5 \%$ efficiency.

At a distance of $5 \mathrm{~m}$ from the source, the intensity produced by the electric field component is evaluated as follows: The intensity at a distance of $5$ meters can be calculated as: Intensity at 5 m

= \frac{5}{4\pi \times 5^2}\left(\frac{W}{m^2}\right)

= \frac{1}{20\pi}\left(\frac{W}{m^2}\right)

Since the intensity due to the electric field component is half of the total intensity: Intensity due to electric field

= \frac{1}{40\pi}\left(\frac{W}{m^2}\right)

Thus, we arrive at our result:

= \frac{1}{40\pi}\left(\frac{W}{m^2}\right)

Q60

Given below are two statements : Statement I : Electromagnetic waves are not deflected by electric and magnetic field. Statement II : The amplitude of electric field and the magnetic field in electromagnetic waves are related to each other as

{E_0} = \sqrt {{{{\mu _0}} \over {{\varepsilon _0}}}} {B_0}

. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are true

B Statement I is true and Statement II is false

C Both Statement I and Statement II are false

D Statement I is false but Statement II is true

Correct Answer

Option B

Solution

Statement I is correct as photon do not carry any charge, hence cannot feel force from either fields.

Statement II is wrong as

E_0=cB_0

{E_0} = {{{B_0}} \over {\sqrt {{\mu _0}{\varepsilon _0}} }}